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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Ex 9.4, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑑𝑦/𝑑π‘₯=(1 βˆ’ cos⁑π‘₯)/(1 + cos⁑π‘₯ ) 𝑑𝑦/𝑑π‘₯ = (1 βˆ’ cos⁑π‘₯)/(1 + cos⁑π‘₯ ) We know that cos 2x = 2cos2 x βˆ’ 1 Putting x = π‘₯/2 cos 2π‘₯/2 = 2 cos2 π‘₯/2 βˆ’ 1 cos x = 2 cos2 π‘₯/2 βˆ’ 1 1 + cos x = 2cos2 π‘₯/2 We know cos 2x = 1 βˆ’ 2sin2 x Putting x = π‘₯/2 cos2 2π‘₯/2 = 1 βˆ’ 2 sin2 π‘₯/2 cos x = 1 βˆ’ 2sin2 π‘₯/2 1 βˆ’ cos x = 2sin2 π‘₯/2 Putting values in equation 𝑑𝑦/𝑑π‘₯ = (2 γ€–sin^2 〗⁑〖π‘₯/2γ€— )/(2 γ€–cos^2 〗⁑〖π‘₯/2γ€— ) 𝑑𝑦 = (γ€–sin^2 〗⁑〖π‘₯/2γ€— )/cos^(2 )⁑〖π‘₯/2γ€— 𝑑π‘₯ 𝑑𝑦 = tan2 π‘₯/2 𝑑π‘₯ Putting tan2 π‘₯/2 = sec2 π‘₯/2 βˆ’ 1 𝑑𝑦 = ("sec2 " π‘₯/2 " βˆ’ 1" )𝑑π‘₯ Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x βˆ’ 1 Putting x = π‘₯/2 tan2 π‘₯/2 = sec2 π‘₯/2 βˆ’ 1 ∫1▒𝑑𝑦 = ∫1β–’(𝑠𝑒𝑐2 π‘₯/2βˆ’1) 𝑑π‘₯ y = ∫1▒〖𝑠𝑒𝑐2 π‘₯/2 𝑑π‘₯βˆ’ ∫1▒𝑑π‘₯γ€— y = 1/(1/2) tan π‘₯/2 βˆ’ x + C y = 2 tan π‘₯/2 βˆ’ x + c y = 2 tan π‘₯/2 βˆ’ x + C y = 𝟐 tan 𝒙/𝟐 βˆ’ x + C is the general solution (As ∫1▒〖𝑠𝑒𝑐2 π‘Žπ‘₯𝑑π‘₯= 1/π‘Žγ€— tan ax + C)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.